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Higher Chemistry
Unit 2
Nature’s Chemistry
What you should know…
• That molecular structure and physical properties of
hydrocarbons are related.
• The names, molecular and structural formula of straight and
branched chain alkanes, alkenes and cycloalkanes.
• How to identify isomers and draw their structural formulae.
• What is meant by saturated and unsaturated carbon compounds
and how they can be distinguished.
• Addition reactions
• The name of the functional group present in alcohols and the
properties of alcohols
• The names, molecular and structural formula of straight and
branched chain alcohols
• The name of the functional group present in carboxylic acids
and the properties of carboxylic acids
• The names, molecular and structural formula of straight and
branched chain carboxylic acids
Organic Chemistry
Originally, chemical compounds were divided into 2 classes:
Inorganic or Organic
Organic compounds were derived from living things. It
was believed that they contained a ‘vital force’ and could
not be made from inorganic compounds (non-living sources).
Organic chemistry is the study of carbon compounds
•Organic molecules may be
as simple as methane, CH4
or as complicated as cholesterol
HO
L.I.
To learn about esters
Section 1 (a)
S.C.
By the end of this lesson you should be able to…
State the name ending present in all esters.
State the name of the functional group present in an ester.
Draw the functional group present in an ester.
State the two types of molecules which contribute to the name of an
ester.
• Draw the structural formula for an ester when given its name.
• Draw the structural formula for an ester when given the name of the
parent alcohol and carboxylic acid.
• Describe the characteristic smell of an ester.
•
•
•
•
Fruit Flavours (a) Esters
An ester can be identified by the name endings ‘-yl –oate’.
Esters contain the carboxylate functional group (–COO-)…
Carboxylate
group
Naming Esters
An ester can be named given the name of the parent alcohol and
carboxylic acid or from shortened and full structural formulae.
The parent alkanol gives the start of the name ending in -yl. The
parent acid gives the second part of the name ending in -oate.
So the ester formed by reacting ethanol with propanoic acid
would be ethyl propanoate…
• Workbook activities
Name of alcohol
Name of carboxylic acid
Name of ester
Ethanol
Propanoic acid
Ethyl propanoate
Propanol
Methanoic acid
Propyl methanoate
Methanol
Butanoic acid
Methyl butanoate
Propanol
Ethanoic acid
Propyl ethanoate
Butanol
Methanoic acid
Butyl methanoate
Ethanol
Butanoic acid
Ethyl butanoate
Full structural formula of ester
1. Methanol
+
Ethanoic acid
Methyl ethanoate
Ethanoic acid
Butyl ethanoate
+
2. Butan-1-ol +
+
L.I.
To learn about making esters
(b)
S.C.
By the end of this lesson you should be able to…
• State the name of the reaction in which esters are formed from
carboxylic acids and alcohols.
• State the definition of a condensation reaction.
• State how an ester linkage can be formed by the condensation
reaction between a hydroxyl group and a carboxyl group, using full
structural formulae.
(b) Making Esters
• In condensation reactions small molecules join together to form a bigger
molecule by the elimination of water.
• Esters are formed by the reaction of a carboxylic acid and an alcohol
• Carboxylic acids contain the functional group carboxyl (-COOH)
• Alcohols contain the functional group hydroxyl (-OH)
Acids and carboxylic acids can react by condensation reactions to form esters.
Alcohol
R
O
alcohol
+ Acid
⇌
O
C R'
H + H O
carboxylic acid
Ester + Water
O
C R' + H2 O
R O
ester
The ester link is formed by the reaction of a hydroxyl group with a carboxyl
group. The reaction is reversible - an equilibrium is set up.
If the water is removed the equilibrium is shifted to the side of the products .
Concentrated sulphuric acid has a high affinity for water and is used to remove
it and shift the equilibrium to the right.
Carry out the experiment to make an ester in the lab (Workcard 1) …
Workbook activity…
 What is the purpose of the wet paper towel?
The paper towel acts as a condenser to prevent
volatile gases escaping (it cools them down).
 Why is the reaction mixture heated in a water bath?
To increase the temperature of the reaction and
make the reaction happen faster. A Bunsen cannot be
used as the reactants are flammable.
 Why is the reaction mixture poured onto sodium
hydrogen carbonate solution?
It was added to the products to neutralise the sulphuric acid
and any excess carboxylic acid.
 Observation:
The product had a strong smell and formed a layer over the water since esters are immicible in water.
Carboxylic Acid
Alcohol
Name of ester
Ethanoic acid
Pentanol
Pentyl ethanoate
Ethanoic acid
Ethanol
Ethyl ethanoate
Methanoic acid
Ethanol
Ethyl methanoate
Methanoic acid
Propanol
Propyl methanoate
Shortened structural
formula of ester
Smell of
ester
CH3COO(CH2)4CH3
banana
CH3COOC2H5
fruity odour
HCOOCH2CH3
rum-like odour
HCOOCH2CH2CH3
plum odour
L.I.
To learn about the uses of esters
S.C.
By the end of this lesson you should be able to…
• Describe the three main uses of esters.
(c)
(c) Uses of esters
Esters are oily liquids with generally very pleasant fruity smells and
have a range of uses.
Many esters are used as flavourings, in perfumes and as solvents.
Workbook activity
Name
Shortened Structural Formula
Odour/Flavour
Pentyl ethanoate
CH3COO(CH2)4CH3
Banana
Octyl ethanoate
CH3COO(CH2)7CH3
Orange
Methyl Butanoate
CH3CH2CH2COOCH3
Pineapple
3-Methylbutyl Butanoate
CH3(CH2)2COO(CH2)2CH(CH3)2
Apple
Propyl ethanoate
CH3COOCH2CH2CH3
Pear
Methyl-1-butyl ethanoate
CH3COOCH(CH3)C4H9
Banana
2-Methylpropyl methanoate
HCOOCH2CH(CH3)CH3
Raspberry
Pentyl butanoate
C3H7COOC(CH2)4CH3
Apricot, Strawberry
Benzyl ethanoate
CH3COOCH2C6H5
Peach, flowers
Ethyl methanoate
HCOOCH2CH3
Rum
Methyl 2-aminobenzoate
C6H4(NH2)COOCH3
Grapes
Benzyl butanoate
C3H7COOCH2C6H5
Cherry
Uses of esters as solvents
Some of the smaller esters are quite volatile and are used as solvents in adhesives, inks and paints –
pentyl ethanoate is used in nail varnish for example.
Ethyl ethanoate is one of a number of solvents used to extract caffeine from coffee and tea.
De-caffeinated products produced with ethyl ethanoate are often described on the packaging as
"naturally decaffeinated" because ethyl ethanoate is a chemical found naturally in many fruits.
Caffeine (C8H10N4O2) is an example of a class of compounds called
alkaloids which are produced by plants.
The name alkaloid means “alkali-like”, where alkali is a base and hence
refers to these basic properties.
Caffeine is more soluble in the
organic solvent ethyl ethanoate
than in water, so we will extract
caffeine into the organic solvent to
separate it from glucose, tannins,
and other water soluble compounds
using a separating funnel.
The ethyl ethanoate portions can
be combined and the ethyl
ethanoate removed by evaporation
to leave the caffeine
Workbook activity
VOC stands for ‘Volatile Organic Compounds’.
Legislation aims to prevent or reduce the direct and indirect effects of emissions of volatile
organic compounds (VOCs) on the environment and human health, by setting emission limits for such
compounds and laying down operating conditions for installations using organic solvents.
Why?
The emissions of volatile organic compounds * (VOCs) in the atmosphere contribute to the
formation of the tropospheric ozone (ozone in the lower atmosphere). Large quantities of this
ozone may be harmful to people, vegetation, forests and crops. Sensitive people may suffer
irritation of the throat and eyes, as well as respiratory difficulties. Tropospheric ozone is also a
greenhouse gas.
L.I.
To learn about the hydrolysis of esters
(d)
S.C.
By the end of this lesson you should be able to…
• Describe how an ester can be hydrolysed using full structural formulae.
• State the definition of a hydrolysis reaction.
• State the name of the products formed by the hydrolysis of an ester.
In hydrolysis reactions large molecules are broken down into smaller molecules by water.
Esters can be hydrolysed to produce the parent ................... .......... and the parent
.................... .
ESTER + WATER
⇌
ACID + ALCOHOL
An alkali is often used to catalyse the reaction and the mixture heated under
reflux.
If sodium hydroxide is used a sodium salt of the acid is formed. If this is reacted with a strong acid
(eg. Hydrochloric acid) the acid can be displaced from the salt
HCOO- Na+
Sodium
methanoate
+
+
HCl
hydrochloric
acid
HCOOH
methanoic
acid
+
+
NaCl
sodium
chloride
Carry out the experiment to hydrolyse ethyl benzoate (Workcard 2)
Mass of filter paper (g)
Mass of filter paper + benzoic acid (g)
Mass of benzoic acid formed (g)
(actual mass)
Use the equation below to calculate the theoretical mass of benzoic acid expected when 0.033
moles (5g) of ethyl benzoate is hydrolysed.
C6H5COOC2H5 + H2O
⇌ C6H5COOH + C2H5OH
Calculate the % Yield using the following equation…
% yield = actual mass
x 100
theoretical mass
L.I.
To learn about edible fats and oils
S.C.
By the end of this lesson you should be able to…
•
•
•
•
Section 2 (a)
State why fats and oils are an essential part of a healthy diet.
Describe the formation and structure of fats and oils.
Explain using full structural formulae how fats and oils are formed.
State the main difference between the structure of saturated and
unsaturated fatty acids.
Natural oils have vegetable and marine (fish) origins
Natural fats have ………………………, …………………… or ………………………… origins
Workbook Activity
Name of fat or oil
Sunflower oil
Linseed oil
Suet
Cod liver oil
Lard
Castor oil
Palm oil
Rapeseed oil
Source
Animal, vegetable
or marine
Name of fat or oil
Source
Sunflower oil
Linseed oil
Suet
Cod liver oil
Lard
Castor oil
Palm oil
Sunflower seeds
Flax seeds
Beef
Cod fish
Pork
Castor plant
Fruits of palm oil
trees
Rapeseed plant
Rapeseed oil
Animal, vegetable
or marine
Vegetable
Vegetable
Animal
Marine
Animal
Vegetable
Vegetable
Vegetable
Fats and oils are an essential part of the diet. They provide the body with energy.
Fats and oils are a more concentrated source of energy than .................................
Carry out the experiment to test the unsaturation of fats and oils (Workcard 3)
 What does unsaturated mean?
 Which sample is the most saturated and which is the most unsaturated?
 This comparison is only approximate. How could the method be improved?
Solids fats tend to be saturated
Liquid oils tend to be unsaturated
Fats and oils are naturally occurring esters of the alcohol glycerol and long chain carboxylic acids.
Long chain carboxylic acids are often called ‘fatty acids’
R1, R2, R3 are long carbon
chains which can be the
same or different
glycerol
Propan-1,2,3-triol
Glycerol is a trihydric alcohol – it has three alcohol groups the correct name is ....................................
1 mole of glycerol - a
3 moles of fatty acids to ..........
Fats and Oils are formed by combination of .....
3
triglyceride. There are ..................
ester linkages in the one molecule.
L.I.
To learn about the melting points of fats and oils
S.C.
By the end of this lesson you should be able to…
Section 2 (b)
• State and explain the difference in melting points of fats and oils, in
terms of structure, close packing and strength of intermolecular bonds.
solid
liquid
• Fats are .................... at room temperature while oils are ...................... .
• The properties of a fat or an oil depend on the fatty acids which are combined with the
glycerol in each triglyceride.
• Most fats and oils in nature are mixtures of triglycerides in which the fatty acid
molecules may or may not be identical.
• Fatty acids are saturated or unsaturated straight chain carboxylic acids. They contain an
even number of carbon atoms and range in size from C4 to C24 carbon atoms, but mainly C16
and C18.
Workbook activity
Complete the table below…
Fatty acid
palmitic acid
stearic acid
linoleic acid
oleic acid
Formula
Saturated or
unsaturated
Fatty acid
palmitic acid
stearic acid
Formula
C16H32O2
C18H36O2
linoleic acid
C18H32O2
oleic acid
C18H34O2
Saturated or
unsaturated
Saturated
Saturated
Unsaturated
Unsaturated
Fat and oil molecules are roughly ‘tunning fork’ in shape with three limbs being hydrocarbon chains.
If the chains are saturated, the molecules pack neatly together even at quite high temperature
Fat – higher melting point
Oil – lower melting point
If the chains contain one or more double bonds the zig-zag chains become more distorted and the
close packing of the molecules is less easy.
 How will the close packing of fat molecules affect the strength of the Van der Waal’s
(intermolecular) bonds?
The more closely the molecules can pack together, the stronger the van der Waals forces.
 What effect will this have on the melting point?
The stronger the van der Waals forces, the more energy is needed to break them and the melting point will
be higher.
L.I.
To learn about the function of proteins
S.C.
By the end of this lesson you should be able to…
• Describe the role of proteins in the body.
Section 3 (a)
growth
repair
We need protein for ................. and ............... . Foods such as meat and fish are rich in protein.
Proteins are the major structural materials of animal tissues.
Proteins are also involved in the maintenance and regulation of life processes.
Proteins can be classified as fibrous or globular.
Fibrous proteins are long and thin and are the major structural materials of animal tissue.
Workbook activity…
Give examples of fibrous proteins…
Keratin found in hair and nails
Collagen and elastin found in connective tissue in the body.
Globular proteins have the spiral chains folded into compact units. Globular proteins are involved in the
maintenance and regulation of life processes and include enzymes and many hormones.
Workbook activity…
Give examples of globular proteins
Albumin found in blood plasma
Haemoglobin found in blood which carries oxygen around the body
Proteins are chemicals containing the element nitrogen.
Carry out the experiment to heat proteins with soda lime (Workcard 4)
Protein
Effect on moist pH paper
When a protein is heated strongly in the presence of soda lime (a mixture of calcium and sodium
hydroxide) an unpleasant smelling alkaline gas is produced which turns moist pH paper blue.
These gases are amines or ammonia - chemicals which contain nitrogen.
Enzymes are proteins, which act as biological catalysts.
They are specific to particular chemical reactions e.g. the enzyme pepsin only catalyses the hydrolysis of
proteins and not any other chemical reaction
Certain sequences of amino acids form a region known as the active site.
The shape of the active site allows specific reactants known as substrates to attach, like a lock and key.
Incorrect substrates are unable to fit the shape of the active site and are not changed.
Enzyme function is therefore related to the molecular shapes of proteins
Carry out the experiment to find out if pH affects enzyme activity (Workcard 5)
1.
‘normal reaction’
Rate
2.
Enzyme reaction
Rate
Temperature
3.
Enzyme reaction
Rate
Temperature
pH
If the temperature rises above a critical level the shape of the enzyme molecule becomes
irreversibly altered although the peptide links remain intact.
This means that the enzyme is unable to function.
denatured
The enzyme is said to be ..............................
.
All proteins, not just enzymes, can be denatured by temperature and pH.
L.I.
To learn about amino acids
Section 3 (b)
S.C.
By the end of this lesson you should be able to…
• State the name of the building blocks from which protein molecules are
formed.
• Describe the structure of amino acid molecules and how they differ to
produce 20 common naturally occurring amino acids.
• Explain why some amino acids are known as essential amino acids.
Amino acids contain an amino group (-NH2) at one end and an acid group (-COOH) at the other end of the
molecule. The majority of amino acids found in proteins are of the type
R represents a carbon side chain which may even contain the elements nitrogen and sulphur.
Name
R (side chain)
Name
R (side chain)
Glycine
-H
Aspartic Acid
-CH2 COOH
Alanine
-CH3
Methionine
-CH2 CH2 S CH3
Valine
-CH CH3 CH3
Phenylalanine
-CH2 C6H5
There are 20 or so amino acids which go into making up protein.
The body cannot make all the amino acids required for body proteins and is dependent on dietary
essential
protein for the supply of certain .............................
amino acids.
L.I.
To learn about amide links
Section 3 (c)
S.C.
By the end of this lesson you should be able to…
• State the type of reaction in which many amino acid molecules can link
together to form a protein molecule.
• Explain using full structural formulae how an amide (peptide) linkage can
be formed by a condensation reaction between amino acids.
• State how the diverse range of proteins needed to fill the different roles
in the body is produced from just 20 amino acids.
Proteins specific to body’s needs are built up within the body.
condensation
Proteins are ........................... polymers formed by combining amino acids to form long chain molecules of
maybe several thousand amino acid units long.
Workbook activity…
The following amino acids react to form a tripeptide.
Draw the structure of the product molecule below.
When amino acids join together an amide link (or peptide link) is formed and water is eliminated.
The amide link is formed by the reaction of an amino group with a carboxyl group.
The structure of a section of protein is based on the constituent amino acids.
The sequence by which the amino acids are joined differs in different protein molecules.
The sequences of amino acids are controlled from information in the nucleus of the cell.
Less complex than the proteins are the peptides.
A tripeptide has 3 amino acid units.
A shorter chain of amino acids, up to 100 units or so is often referred to as a polypeptide.
L.I.
To learn about the hydrolysis of proteins
Section 3 (d)
S.C.
By the end of this lesson you should be able to…
• State and explain how proteins can be hydrolysed (digested) to produce
amino acids.
• Draw the full structural formulae of the amino acids obtained from the
hydrolysis of a given section of a protein.
• Explain how the amino acids present in a sample of hydrolysed protein can
be analysed and identified.
During digestion enzymes hydrolyse the proteins in food to amino acids. These can then pass through the
gut wall into the blood stream.
In the stomach, the enzyme pepsin starts the digestion of protein by hydrolysing to smaller polypeptides.
Further enzymes in the gut continue the hydrolysis to the constituent amino acids.
In hydrolysis large molecules are broken down into smaller molecules by water.
PROTEIN + WATER → AMINO ACIDS
Workbook activity
Draw the product molecules as this tripeptide molecule is hydrolysed.
In the lab a protein can be hydrolysed back to its constituent amino acids by refluxing with concentrated
hydrochloric acid for several hours.
Amino acids can be identified by the use of paper (or thin layer) chromatography.
A piece of chromatography paper is spotted with some amino acids suspected as being present and also with
the hydrolysed protein.
By comparing the position of the spots of the known amino acids with
that of the hydrolysed protein, the amino acids in the protein can be
identified.
L.I.
To learn about flavour in food
Section 4 (a)
S.C.
By the end of this lesson you should be able to…
• Explain why volatile molecules are important in flavour.
• Identify the functional groups present in flavour molecules and explain
whether they are likely to be water soluble or oil soluble.
• Make predictions about the boiling point (and volatility) of flavour
molecules based on molecular size and the functional groups present.
Many of the flavours in foods are due to the presence of volatile molecules. Normally these molecules
are trapped in the cell. However, during cooking the cell walls can be broken by moisture within the cell
evaporating and rupturing the walls. Also chemicals can damage the walls which are made of a structural
carbohydrate called cellulose.
Volatile flavour molecules can then leave the food and enter the air. Also some may dissolve in the
cooking liquid and be lost. So the method and time of cooking is important in preserving flavours.
Flavour molecules can be water-soluble or oil/fat-soluble. You can preserve the flavour by using water for
cooking foods that contain oil/fat-soluble flavours such as green beans and broccoli, and using oil/fat to
cook foods that are richer in water-soluble flavours such as asparagus. Overcooking food using either
method will cause more flavour to be lost.
To recognise flavours we use our sense of smell and sense of taste. Try eating flavoured food without
looking at it, while pinching your nose. It is not as simple are you might think. There are at least five basic
taste qualities: sweet, sour, bitter, salty, and umami. (Umami, or savoury, is the taste we comes from
glutamate, found in chicken broth, meat extracts, and some cheeses.) Our nose can detect more than
10,000 different smells. If there is hydrogen bonding between the molecules responsible to smell, this
will reduce the volatility of the molecule.
To complicate matters some chemical reactions that take place during cooking can result in new,
desirable flavours. The chemistry here is complicated but a simple example to illustrate that more
desirable flavours appear happens when bread is toasted. Bread contains a carbohydrate and its
carbonyl group (>C=O) reacts with amino groups (-NH2) in protein to make this desirable smell.
The functional groups present in flavour molecules will give an indication whether they are likely to
be water or oil soluble. The size and functional groups present can be taken into account in
predicting their relative boiling point and hence probable volatility. The important feature here is
whether the functional groups present can hydrogen bond with water, making them water
soluble, or whether they are unable to hydrogen bond with water in which case they will be fator oil-soluble.
Some functional groups are included in the table below for reference. However, the presence of a
particular functional group is no guarantee of the substance being water- or oil-soluble. e.g. a long
hydrocarbon chain (which is non-polar) with a single polar group would be unlikely to be able to
hydrogen bond with water and would make the molecule more likely to be oil-soluble.
Functional group name
Functional group structure
Can it form hydrogen bonds?
Water- or oil-soluble
Alkane
C-C
No
oil-soluble
Alkene
C=C
No
oil-soluble
Alkyne
C=C
No
oil-soluble
Hydroxyl
-O-H
Yes
water-soluble
Carbonyl
>C=O
Yes
water-soluble
Carboxylic acid
-COOH or
Yes
water-soluble
Aldehyde
-CHO or
Yes
water-soluble
Amino
-NH2 or
Yes
water-soluble
Phenyl
C6H5-
No
oil-soluble
Amide
-CONH- or
Yes
water-soluble
Ester
-COO- or
Yes
water-soluble
As a general rule, if the molecule contains only C and H atoms is will be non-polar, while if it also
contains O or and N atoms it is more likely to be polar. However, the non-polarity of long hydrocarbon
chains/rings can cancel the effect of a single polar group and make the molecule behave as non-polar.
Workbook activity…
The molecular structures below represent the formulae of some common flavours. Examine
them and complete the table to show if they are likely to be water or oil soluble.
Molecular Name
Flavour
Nona-2,6-dienal
Cucumber
4-hydroxy-2,5dimethylfuran-3-one
Propyl pentanoate
Strawberry
Limonene
Orange
1-methyl-1-butylethanoate
Banana
3,7-dimethyloct-2,6-dienal
Lemons
2-methoxy-3-isobutylpyrazine
Vanillin
Green peppers
Methyl anthranilate
Pineapple
Pineapple
Vanilla
Water soluble or
Oil soluble
Functional groups present
in molecule
Molecular Name
Flavour
Nona-2,6-dienal
Cucumber
4-hydroxy-2,5dimethylfuran-3-one
Propyl pentanoate
Strawberry
Limonene
Orange
1-methyl-1-butylethanoate
Banana
3,7-dimethyloct-2,6-dienal
Lemons
2-methoxy-3-isobutylpyrazine
Vanillin
Green peppers
Methyl anthranilate
Pineapple
Pineapple
Vanilla
Water soluble or
Oil soluble
Functional groups present
in molecule
In general, molecules with a molecular mass of under 300 are likely to be volatile whereas much larger
molecules are not likely to be volatile. The type of bonding present will also have a significant influence
on how volatile the molecule is likely to be. If hydrogen bonding is not present between molecules then
these molecules will be more volatile due to the weak intermolecular forces present.
Watch Heston’s cooking video clips
L.I.
To learn about changes in protein structure upon heating Section 4 (b)
S.C.
By the end of this lesson you should be able to…
• Explain the importance of intermolecular bonding in protein structure.
• State the difference between fibrous and globular proteins.
• State the changes that take place on heating proteins and relate this to
the texture of foods such as eggs and meat.
Proteins are complex molecules made from long, amino acid chains which are also branched. The chains are
held together by intermolecular bonding between the side chains of the constituent amino acids. Hydrogen
bonds occur between the amide links and between other groups present in the molecule. Proteins form
three-dimensional structures which are either sheets, spirals or coils.
When proteins are heated, during cooking, these intermolecular bonds are broken allowing the proteins to
change shape (denature). These changes alter the texture of foods.
For example, egg whites contain many molecules of a globular protein called albumen. When an egg is boiled
or fried, the protein structure is irreversibly changed and a solid is made. During cooking, the protein is
denatured and the protein chains unwind and, as they can now form intermolecular bonds with
neighbouring albumen molecules, a network of interconnected proteins forms causing the egg white to
solidify.
Collagen molecule
The structure of meat changes when cooked. Different temperatures are required for cooking meats with
different levels of connective tissue. Joints containing a lot of connective tissue become tender if cooked
at over 60oC as the collagen forming the tough connective tissue, denatures. The tender lean meat found
in cuts such as fillet steaks have less connective tissue and should not be cooked at too high a
temperature, because in this case the protein molecules start to bunch together resulting in the meat
becoming tougher.
Workbook activity…
Carry out research and write a short note on the chemistry of browning foods. Use the ipads to help you
answer the following questions...
 What are Maillard reactions?
 What reaction takes place during caramelisation?
 Why do apples brown?
L.I.
To learn about the oxidation of alcohols
S.C.
By the end of this lesson you should be able to…
Section 5 (a)
• Draw the structural formula of branched chain alcohols when given the name.
• State the name of branched chain alcohols when given the structural formula.
• State the molecular formula of branched chain alcohols when given the name
or structural formula.
• State how to classify alcohols as primary, secondary or tertiary.
• State the product formed when primary alcohols are oxidised.
• State the product formed when secondary alcohols are oxidised.
• State what happens when attempting to oxidise a tertiary alcohol.
• State what is meant by oxidation in terms of the oxygen to hydrogen ratio.
• State the name of common oxidising agents used in the lab and describe the
colour changes which occur.
• State the ion equations associated with the oxidation of primary and
secondary alcohols.
• State what is meant by the terms ‘dihydric’ and ‘diol’.
• State what is meant by the terms ‘trihydric’ and ‘triol’.
•Draw the structural formulae of dihydric and trihydric alcohols when
given the name.
•State the effect of hydrogen bonding on the properties of diols and
triols.
Alcohols can be classified according to the position of the –OH group.
• In a primary (1°) alcohol, the carbon which carries the -OH group is
only attached to one alkyl group (chain of carbon atoms). Methanol,
CH3OH, is counted as a primary alcohol even though there are no alkyl
groups attached to the carbon with the -OH group on it.
• In a secondary (2°) alcohol, the carbon with the -OH group attached is
joined directly to two alkyl groups, which may be the same or
different.
• In a tertiary (3°) alcohol, the carbon atom holding the -OH group is
attached directly to three alkyl groups, which may be any combination
of same or different.
Structural formula
Name
ethanol
2-methylpropan-2-ol
1o 2o 3o
Alcohols burn in oxygen and air to produce carbon dioxide and water.
This is complete oxidation (combustion).
C2H5OH + 3O2
2CO2 + 3H2O
Less vigorous oxidising conditions can be provided by reacting an alcohol with an oxidising agent.
Your teacher will demonstrate the oxidation of alcohols using hot copper oxide (Workcard 6a)
Carry out the experiment using potassium dichromate (Workcard 6b)
Oxidising alcohol using hot copper oxide
The alcohol vapour is passed over hot CuO. The copper oxide is reduced to copper metal and the alcohol is
oxidised.
Ethanol + copper oxide
C2H5OH + CuO
ethanoic acid + copper + water
CH3COOH + Cu + H2O
Oxidising alcohols using potassium dichromate
The dichromate ion is reduced to the Cr3+ ion and the alcohol is oxidised.
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Primary alcohols are oxidised, first to aldehydes and then to carboxylic acids.
Secondary alcohols oxidise to ketones.
Tertiary alcohols are resistant to oxidation
Oxygen to hydrogen ratio
For carbon compounds, oxidation results in an increase in the oxygen to hydrogen ratio and
reduction results in a decrease in the oxygen to hydrogen.
These are different definitions from those met previously and only apply in carbon chemistry.
eg. C2H5OH
Oxygen : hydrogen ratio
(O:H)
1:6
oxidation
CH3CHO
1:4
Oxidation of ethanol (C2H5OH) to ethanal (CH3CHO) has increased the O:H ratio.
Ethanal can also be reduced back to ethanol and this would be a decrease of the ratio.
Workbook activity…
Name
Formula
C2H5OH
Ethanal
Ethanoic acid
CH3CHO
CH3COOH
Oxygen:Hydrogen Ratio
Name
Formula
Oxygen:Hydrogen ratio
Ethanol
C2H5OH
1:6
Ethanal
CH3CHO
1:4
Ethanoic acid
CH3COOH
2:4
= 1:2
Now complete the workbook activity by calculating the O:H ratio for each of the compounds given.
Remember to draw the structural formulae of each compound too.
• More than one hydroxyl group
• There are some alcohols that have more than one hydroxyl group in the molecule. For
example, glycol, which is used as the antifreeze in car radiators, and glycerol, which you may
have eaten as an ingredient in soft ice cream.
• The alcohol used in ordinary car antifreeze has two hydroxyl groups and for this reason is
known as a dihydric alcohol or diol. Its common name is glycol, but this says little about its
structure. The systematic name is ethane-1,2-diol...
Two numbers are needed in the name to describe the positions of the two hydroxyl groups.
Notice that ‘di’ now appears infront of ‘ol’ (‘di’ = 2, ‘ol’ = hydroxyl group).
With two carbon atoms in the molecule, the systematic name is based on ethane. In this case the final ‘e’ of
ethane is not dropped in the sysetmatic name because it is not followed by a vowel.
The alcohol commonly known as glycerol is found in a variety of foods. It has the systematic name propane1,2,3-triol.
This molecule has three hydroxyl groups and as a result is known as a trihydric alcohol or triol.
• As the number of hydroxyl groups in a molecule increases,
the number of hydrogen bonds between molecules also
increases. This will mean that a greater amount of energy is
needed to overcome these intermolecular forces of
attraction.
• This explains why propane-1,2,3-triol has a higher boiling
point than ethane-1,2-diol.
L.I.
To learn about aldehydes and ketones
S.C.
By the end of this lesson you should be able to…
Section 5 (b)
• State the name of the functional group present in aldehydes and ketones.
• State the name ending present in aldehydes.
• State the name ending present in ketones.
• State the name of straight chain and branched chain aldehydes when given the
structural formula.
• State the name of straight chain and branched chain ketones when given the
structural formula.
• Draw the structural formula of aldehydes and ketones when given the names.
• State the molecular formula of aldehydes and ketones when given the names.
• State and explain the results of a chemical test used to distinguish between
aldehydes and ketones.
• State the name of the products formed when aldehydes (and ketones) are oxidised.
• State the effect of oxidising aldehydes using chemicals such as Benedict’s
solution, Tollen’s reagent and acidified potassium dichromate solution.
C
Both aldehydes and ketones contain the carbonyl group.
O
In aldehydes a hydrogen atom is bonded to the carbonyl group but in ketones the carbonyl group is
always flanked by carbon atoms:
O
C H
aldehyde
O
C
C C
ketone
This structural difference accounts for the fact that aldehydes can undergo mild oxidation to form
carboxylic acids but ketones resist oxidation.
Oxidising agents can therefore be used to distinguish between aldehydes and ketones.
Carry out the experiment to oxidise aldehydes and ketones (Workcard 7)
Aldehydes but not ketones can be oxidised by a number of oxidising agents, including
Benedict's solution, Tollen’s reagent and acidified dichromate to carboxylic acids.
Benedicts solution
Tollen’s reagent
with compound X
Colour change from
blue to orange-red
with compound Y
No reaction
A solid silver
precipitate forms
(the silver mirror
effect)
No reaction
Compound X must be ………………………………
Compound Y must be ………………………………..
Acidified
dichromate
(H+/Cr2O72-)
Colour change from
orange to green
No reaction
Aldehydes contain the functional group (-CHO)…
Aldehydes can be identified from the '-al' name ending.
Ketones contain the functional group…
Ketones can be identified from the '-one' name ending.
Aldehydes have a hydrogen atom joined to the carbonyl group whereas ketones have two carbon atoms
joined to the carbonyl group.
Naming of alkanones is almost exactly the same as for alkanols. e.g. butanone. For longer chain ketones the
position of the carbonyl group must be given.
Workbook activities…
1. Draw the full structural formulae and shortened structural formulae for the compounds
listed.
2. Draw the full structural formulae and name each of the compounds listed.
•
•
•
•
•
L.I.
To learn about antioxidants
S.C.
By the end of this lesson you should be able to…
Section 5 (c)
State how food containing edible oils turns rancid.
State how to prevent food from turning rancid.
State the ion-electron equations for the oxidation of antioxidants.
State the name of carboxylic acids when given the structural formula.
Draw the structural formula of branched chain carboxylic acids when given the
name.
• State the molecular formula of branched chain carboxylic acids when given the
names or structural formula.
• State the products formed when carboxylic acids undergo reduction reactions.
• State the name of the product formed when carboxylic acids react with bases.
Oxygen reacts with edible fats and oils in food giving the food a rancid flavour.
For example, butter, a mainly saturated fat, can undergo hydrolysis releasing small quantities of
volatile, unpleasant smelling butanoic acid which is responsible for a rancid smell.
The reaction with oils is more complex and appears to affect unsaturated fatty acids in the oil. The
point of attack is at the C=C double bonds and volatile, smaller molecules such as aldehydes and
ketones are released which cause the rancid smell.
Oxidation reactions can produce free radicals. A free radical is a highly reactive species containing an
unpaired electron. Free radicals can damage food by removal of an electron.
Antioxidant molecules ‘mop up’ free radicals to protect the foodstuff.
• The antioxidant molecule donates an electron to the potentially damaging free radical.
• A stable electron pair is formed, stabilising the free radical.
• The antioxidant itself becomes oxidised (loses an electron).
• Antioxidants are molecules which will prevent oxidation reactions taking place,
antioxidants are reducing agents. By undergoing oxidation, the antioxidant provides
electrons to prevent the oxidation of fats/oils in food.
• In crisp manufacture, potatoes are typically fried under an atmosphere of steam and
packaged under nitrogen to help reduce oxidation.
• Ascorbic acid (or vitamin C), found in many fruits is an antioxidant.
When it behaves as an antioxidant, two of the hydroxyl groups become changed into carbonyl groups as
found in ketones. The product has two hydrogen atoms less than the ascorbic acid and is know as
dehydroascorbic acid.
It can be seen as an oxidation because there is a increase in the O:H ratio and also that electrons
are released in the reaction.
C6H8O6
C6H6O6 + 2H+ + 2e-
When apples are cut, they go brown because of oxidation by the air.
A few drops of lemon juice will prevent this oxidation, though the lemon juice is not an antioxidant.
HOC6H4OH
OC6H4O + 2H+ + 2e-
Citric acid and benzoic acid is used in many foods such as jam to prevent oxidation by helping
other antioxidants do their work.
Antioxidants in action
Oxidation occurs when the
apple is left exposed to air
The apple is protected when
dipped in orange juice
containing the antioxidant
vitamin C
The table shows some typical antioxidants:
Antioxidant
E-number
Typical foods
Beers, cut fruits, jams, dried potato. Helps to prevent cut and
pulped foods from going brown by preventing oxidation reactions
that cause the discolouration. Can be added to foods, such as
potato, to replace vitamin C lost in processing.
Oils, meat pies. Obtained from soya beans and maize. Reduces
oxidation of fatty acids and some vitamins.
Ascorbic acid
(vitamin C)
E300
Tocopherols
E306
Butylated
hydroxyanisole
(BHA)
E320
Oils, margarine, cheese, crisps. Helps to prevent the reactions that
break down fats and cause the food to go rancid .
E330
Jam, tinned fruit, biscuits, alcoholic drinks, cheese, dried soup.
Naturally-occuring in citrus fruits like lemons. Helps to increase the
anti-oxidant effects of other substances. Helps to reduce the
reactions that can discolour fruits. May also be used to regulate pH
in jams and jellies.
Citric acid
http://www.understandingfoodadditives.org
Carry out the SSERC experiments on antioxidants (Workcard 8)
Now carry out the ‘active talk ‘activity with your group
Carboxylic acids
Vinegar is a dilute solution of ethanoic acid which is commonly used in food preservation. Ethanoic
acid is a member of the carboxylic acid family. As a homologous series, carboxylic acids all share
the same functional group and fit the same general formula (CnH2n+1COOH).
Carboxyl
functional group
It is necessary to be able to name and draw structural formulae and molecular
formulae for carboxylic acid molecules. This can be done by using the principles we
have used before...
• Carry out the workbook activities on naming carboxylic acids and drawing full and shortened
structural formulae.
Reactions of carboxylic acids
Carboxylic acids can be reduced in stages back to the original parent aldehyde and alcohol using a
reducing agent such as lithium aluminium hydride.
For example...
Ethanoic acid
(Carboxylic acid)
Ethanal
(Aldehyde)
Ethanol
(Primary Alcohol)
Workbook activity
Like ordinary acids carboxylic acids can react with bases to form salts.
For example...
Ethanoic acid + Sodium hydroxide
+
NaOH
Workbook activity
Sodium ethanoate + Water
+ H2O
* Vitamin C tablets experiment *
L.I.
To learn about soaps and emulsions
Section 6 (a) making soap
S.C.
By the end of this lesson you should be able to…
• Explain how soaps are produced by alkaline hydrolysis of fats and oils.
Soaps are produced by the alkaline hydrolysis of fats and oils.
Fats and oils are esters.
The hydrolysis of fats and oils produces fatty acids and glycerol (propane-1,2,3-triol) in the ratio of three
moles of fatty acid to one mole of glycerol.
In the presence of an alkali, a water soluble ionic salt of the carboxylic acid is formed.
These salts are the important ingredients of soap - the ones that do the cleaning.
Watch the animation of making a soap on
http://www.educationscotland.gov.uk/highersciences/chemistry/animations/soapformation.asp
L.I.
To learn about soaps and emulsions
Section 6 (b) cleansing action of soaps
S.C.
By the end of this lesson you should be able to…
State what is meant by a hydrophobic tail in a soap molecule.
• State what is meant by a hydrophilic head in a soap molecule.
• Describe the cleansing action of soaps in terms of the structure of the soap
molecule.
•
Watch the animation to investigate the cleansing action of soaps...
http://www.educationscotland.gov.uk/highersciences/chemistry/animations/cleansingsoap.asp
Cleaning with water alone has little effect if the stains consist of non-polar substances, such as
grease and sweat.
Soaps and detergents are ‘emulsifying reagents’. These are simply chemicals which can make oil and
water become permanently mixed to produce a stable emulsion.
Soaps and detergents do this job due to the structure of the molecules:
Soap molecules have a long non polar hydrocarbon chain ‘tail’ which is readily soluble in non polar
(hydrophobic) compounds and a polar ionic carboxylate ‘head’ which is water soluble.
During cleaning, the hydrophobic tails dissolve in the droplet of grease, whilst the hydrophilic
heads face out into the surrounding water, resulting in ball like structures. The following
diagrams show how this works.
The dirt or grease is held inside the ball and suspended in the water.
L.I.
To learn about soaps and emulsions
S.C.
By the end of this lesson you should be able to…
• State
the meaning of the term emulsion.
• Explain why emulsifiers are added to food.
• State how emulsifiers are made.
• Describe how emulsifiers work.
Section 6 (c) emulsions in food
An emulsion contains small droplets of one liquid dispersed in an another liquid.
Emulsions in food are mixtures of oil and water.
To prevent oil and water components separating into layers, a soap-like molecule known as an
emulsifier is added.
Emulsifiers for use in food are commonly made by reacting edible oils with glycerol to form
molecules in which either one or two fatty acid groups are linked to a glycerol backbone rather
than the three normally found in edible oils.
The one or two hydroxyl groups present in these molecules are hydrophilic whilst the fatty acid
chains are hydrophobic.
The presence of this emulsifier is shown on packaging by E-numbers, E471 and is one of the
most common on food packaging.
Many foods contain emulsions. Mayonnaise is mainly a mixture of vegetable oil and
water/vinegar/lemon juice. Egg yolk or a synthetic emulsifier (xanthan gum) can be used to keep
the normally immiscible liquid evenly mixed. Without the emulsifier the two liquids would
separate and would not appear appetising.
Emulsifiers are added to a very large range of different foods including sauces, bread, biscuits,
ice cream, low fat spread and even dried pastas where they help to prevent pasta pieces sticking
to each other during cooking.
Watch the animation on emulsions and emulsifiers
http://www.educationscotland.gov.uk/highersciences/chemistry/animations/emulsions.asp
Emulsifiers
Mayonnaise contains oil and water. The emulsifier keeps these mixed and without it the oil
and water separate.
L.I.
To learn about fragrances
S.C.
By the end of this lesson you should be able to…
• State
Section 7 (a) essential oils
what is meant by an essential oil.
•State the uses, properties and products of essential oils.
•Describe how essential oils can be extracted from plant material.
•State the key component in many essential oils.
Essential oils are hydrophobic liquids containing mixtures of volatile aroma compounds that
evaporate easily in air giving distinctive fragrances.
The oils have the aroma of the plant from which they are extracted. They include lavender,
peppermint, orange, lemon, and eucalyptus oils.
Every essential oil contains a mixture of organic compounds rather than just one pure
compound, but it is often the aromatic molecules that provide the distinct fragrances.
Essential oils are extracted from plant material, the difficulty is obtaining the liquid before it
evaporates.
Steam Distillation
In this process the volatility of the oil is important so it is carried by the steam. More water
can be added from the funnel.
The essential oil collects as a mixture of oil and water, but as the two do not mix, they are easily
separated.
In the laboratory extraction can be achieved with the apparatus shown....
Essential oils are used in perfumes, cosmetics, soaps and other products, for flavoring food and
drink, and for adding scents to incense and household cleaning products.
Essential oil can contain esters, alcohols, aldehydes and ketones.
An important component in essential oils are known as terpenes.
Modern uses
Cosmetics
Cleaning
Flavours
Dentistry
Essential oils
Adhesives
Insect
repellents
Medical
Perfumes
Terpenes are unsaturated compounds formed by joining together isoprene (2-methylbuta-1,3-diene)
units.
5 carbon molecule….
(Looks
like a
horse….)
The basic molecular formulae of terpenes are multiples of that, (C5H8)n where n is the number
of linked isoprene units. This is called the isoprene rule.
The isoprene units may be linked together "head to tail" to form linear chains or they may be
arranged to form rings. One can consider the isoprene unit as one of nature's common building
blocks
Complete the workbook activity to highlight the isoprene units in each molecule.
Terpenes are components in a wide variety of fruit and floral flavours and aromas.
Terpenes can be oxidised within plants producing some of the compounds responsible for
the distinctive aroma of spices.
Limonene – a cyclic terpene
CH2
H3C
C
CH
H2C
CH2
H2C
CH
C
CH3
Limonene
(skin of citrus fruits)
Menthol – a cyclic terpenoid
CH3
H3C
This terpene has been
oxidised to a terpenoid
CH
CH
OH
H2C
CH
H2C
CH2
CH
CH3
Menthol
(peppermint)
a-Selinene – a cyclic terpene
CH2
H2C
H2C
CH
C
CH3
3 isoprene units
CH2
C
CH2
C
CH2
CH2
C
H
CH3
a-Selinene
CH2
15 carbon atoms
β-carotene – a linear terpene
H3C
H3C
C
C
H2C
C
CH2
CH
C
CH
H2C
CH
CH3
8 isoprene units
40 carbon atoms
C
CH2
C
CH2
CH3
CH3
CH3
CH2
CH
CH
CH
CH
-carotene
C
CH3
CH
CH
CH
CH
CH
C
CH
C
CH3
C
CH
H3C
CH3
Find the horse…
L.I.
To learn about skin care products
S.C.
By the end of this lesson you should be able to…
• State
Section 8 (a) effect of UV light
the effect of ultraviolet radiation on molecules.
•State the effect of ultraviolet radiation on the human body.
•Describe how sunblock works.
Image of the sun
taken with ultraviolet
imaging telescope
Ultraviolet radiation (UV) is a high-energy form of light, present in sunlight.
Exposure to UV light can result in molecules gaining sufficient energy for bonds to be
broken. This is the process responsible for sunburn and also contributes to ageing of the
skin. Sun-block products prevent UV light reaching the skin.
UV light is divided into UVA, UVB and UVC light. UVA is the highest energy and causes most
damage. UVC light does not penetrate the atmosphere and so causes no problems.
UVA and UVB cause wrinkles by breaking down collagen, creating substances called free
radicals that inhibit the natural the repair of the skin.
UVB light, but not UVA light is stopped by glass. You may notice labels on sunglasses to
indicate their effectiveness.
UV Photography reveals the effects of "photoaging", or ageing of skin caused by light.
Photoageing refers to the damage that is done to the skin from prolonged exposure to UV
radiation, over a person's lifetime.
Most of the skin changes that occur as we get older are accelerated by sun exposure.
Examples of skin changes from photoaging include dark spots, wrinkles, leathery skin and skin
cancer (malignant melanoma).
There’s no such thing as a healthy tan
Skin cancer – malignant melanoma
In the UK, 2,000 people a year die from malignant melanoma, and the number is increasing.
Photoageing caused by UVA
Taxi driver
Exposed to sun
(through car window)
Not exposed to sun
The effects of UV - ageing of skin
How to protect yourself-Sunscreen
•
Sunscreen works by combining organic and inorganic active
ingredients.
•
Inorganic ingredients like zinc oxide or titanium oxide reflect or
scatter ultraviolet (UV) radiation.
•
Organic ingredients absorb UV radiation, dissipating it as heat.
L.I.
To learn about skin care products
Section 8 (b) Free Radical Reactions
S.C.
By the end of this lesson you should be able to…
State what is meant by a free radical.
• Describe the reactivity of a free radical.
• Describe how free radicals are formed.
• State what is meant by the terms initiation, propagation and termination.
•
When UV light breaks bonds free radicals are formed.
Free radicals have unpaired electrons and, as a result, are highly reactive.
Free radical chain reactions include the following steps: initiation, propagation and termination.
Hydrogen and chlorine
video clip
Hydrogen reacts with explosively with chlorine in the presence of U.V. light. The reaction can be shown as
follows:
H2(g) + Cl2(g) → 2HCl(g)
The presence of acid HCl in the product can be shown with moist pH paper.
The reaction follows a free radical chain reaction, initiated by U.V. light. For convenience, the reaction can
be split into three stages.
1) Initiation
U.V. light provides the energy for the homolytic fission of halogen into reactive halogen atoms or free
radicals (atoms with an unpaired electron).
.
Cl2(g) → Cl (g) +
.Cl(g)
2) Propagation
In this stage, free radicals collide with other species but the number of free radicals is maintained (hence
the term propagation).
.Cl → H.(g) + HCl(g)
H.(g) + Cl2(g) → HCl(g) + Cl. (g)
H2(g) +
These reactions continue until reactants are used up, or until free radicals are used up by collision with each
other.
3) Termination
In this stage, free radicals are used up by collision with each other.
H.(g) +
.Cl(g) → HCl(g)
H.(g) + .H(g) → H2(g)
Cl.(g) + .Cl(g) → Cl2(g)
Free radical Substitution Methane
Another free radical reaction takes place when a halogen is substituted into an alkane in the presence of UV
light. This reaction is not explosive and results in the decolourisation of bromine.
Alkanes react with bromine in the presence of U.V. light, though the reaction with bromine is slow. The
reaction can be shown as follows:
CH4(g) + Br2(g) → CH3Br(g) + HBr(g)
The presence of acid HBr in the product can be shown with moist pH paper.
However, the reaction does not end here and further substitution can occur with hydrogen atoms
progressively replaced by halogen atoms.
The slow substitution reaction follows a free radical chain reaction, initiated by U.V. light. Again, the
reaction can be split into three stages…
1) Initiation
U.V. light provides the energy for the homolytic fission of halogen into reactive halogen atoms or
free radicals (atoms or molecular fragments with an unpaired electron).
Br2(g) → Br.(g) + .Br(g)
2) Propagation
In this stage, free radicals collide with other species but the number of free radicals is
maintained (hence the term propagation).
CH3-H(g) + .Br → CH3.(g) + HBr(g)
CH3.(g) + Br2(g) → CH3-Br(g) + Br. (g)
These reactions continue until reactants are used up, or until free radicals are used up by collision
with each other.
3) Termination
In this stage, free radicals are used up by collision with each other.
Br.(g) +
CH3.(g) +
CH3.(g) +
.Br(g) → Br2(g)
.Br(g) → CH3-Br(g)
.CH3(g) → CH3-CH3(g)
The product of the last equation is ethane. However, to minimise the range of possible products, an excess
of the original alkane is used and the products separated from the excess alkane by distillation.
Evidence to support this mechanism



The reaction is initiated by U.V. light and, once started, can continue in the dark.
Other substitution products are made such as CH2Br2, CHBr3, CBr4 together with longer alkanes (and
smaller amounts of substitution products of these alkanes.
However, these other substitution products can be minimised by using an excess of the original alkane to
try to ensure collision of the relatively small number of free radicals produced by sunlight quickly uses up
the bromine.
L.I.
To learn about skin care products
Section 8 (c) Free Radical Scavengers
S.C.
By the end of this lesson you should be able to…
State the effect that free radicals have on the body.
• Describe how free radical scavengers work.
• State examples of natural free radical scavengers.
•
Many cosmetic products contain free radical scavengers.
These are molecules which can react with free radicals to form stable molecules and prevent
chain reactions.
Melatonin and Vitamin E are examples of natural free radical scavengers.
Melatonin
As UV light can cause wrinkling of skin, some skin-care products claim to contain chemicals
which prevent wrinkling. These are claimed to be anti-ageing creams.
As free radical scavengers form stable molecules, they help to stop the free radical chain reactions
that cause the skin to form wrinkles.
Free radicals remove electrons from skin cells and damage them and wrinkles start to develop.
Watch a selection of adverts for anti-ageing products. Identify the scientific
basis of the claims.
Permanent
benefits?
Do they work?
There is a range of antioxidants used in anti-wrinkle creams, and some are better at penetrating the skin than others. The
antioxidants used in skin care are derived from Vitamins A, E and C
The derivatives of Vitamins A (retinol) and E combat free radicals. Vitamin C is used in the construction of collagen. Other
antioxidants work by exfoliating the dead skin on the surface to reveal newer, younger-looking skin underneath. Still
others create a barrier to prevent moisture loss fromClaims for retinol derivatives say it can reduce the appearance of
lines and reduce skin roughness, and blotchiness. There is some research that says retinol can increase the thickness of
the epidermis. But as the molecules are large, they can't fit though the skin unless combined with substances that make
the holes in the lipid matrix bigger.
Vitamin E is the most widely used ingredient in skin care products, used for its moisturising and antioxidant properties.
Free radical scavengers are also added to food products to slow down the rate of oxidation and also to
plastics as without them they can break down under the action of UV light and oxygen.
Carry out the experiment to make your own lip balm (Workcard 11).
Revision Questions…
Butanoic acid has a powerful, unpleasant odour. It is
found in rancid butter, Parmesan cheese and vomit.
Carboxyl group
Can you identify and name the functional group
present in butanoic acid?
Aqueous solutions of methanal are commonly used in
embalming to preserve human or animal remains.
Carbonyl group
Can you identify and name the functional group present
in methanal?
Is methanal an aldehyde or a ketone?
Aldehyde
The molecule below is found in the disinfectant Dettol,
which we instantly recognise by its distinctive smell.
Dettol (chloroxylenol) helps us fight unwanted
bacteria.
Hydroxyl
group
Can you identify and name the functional group
present in Dettol?
4-formamidobenzoic acid is used in pharmaceutical
compositions.
Carboxyl
group
Amide group
Can you identify and name two functional groups
present on the 4-formamidobenzoic acid molecule?
Cinnamon is a tasty spice used to flavour biscuits,
cakes and pies. Cinnamon also has medicinal
properties.
Carbonyl
group
Carbon-tocarbon double
bond
Can you identify and name two functional groups in
cinnamon?
Is cinnamon an aldehyde or a ketone? Aldehyde
What is the strongest type of intermolecular force of
attraction between cinnamon molecules?
(Hint: Think about the bond polarity of the functional
groups present!)
Methyl anthranilate occurs naturally in grapes and
is used as grape flavouring in drinks and chewing
gum.
Ester link
Amino group
Can you identify and name the two functional
groups present in methyl anthranilate?
Vitamin C is needed in your diet for the growth and
repair of tissues in all parts of your body.
Ester link
Carbon-tocarbon double
bond
Hydroxyl
group
Can you identify and name three different types of
functional group on the structure of vitamin C?
Which is the strongest type of intermolecular force
of attraction between molecules of vitamin C?
Is vitamin C a polar or a non-polar molecule?
Is vitamin C soluble in water or in hexane?
(Hint: Think about which functional groups are present!)
Glucose is a simple sugar that is used as an energy
source by many living organisms.
Is glucose soluble in water or in hexane? Justify your
answer with reference to the functional groups present.
2-methylpropane is a branched hydrocarbon that is
used as a refrigerant.
Is 2-methylpropane soluble in water or in hexane?
Justify your answer with reference to the structure.